If vertices of a triangle are A(1,- 1, Z), B(2,0,- 1) and C(0,2,1), then the area of a triangle is

Now,Δxy=121−11201021=12|1(0−2)+1(2−0)+1(4−0)|=12|−2+2+4|=2Δyz=12−1210−11211=12|[−1(−1−1)−2(0−2)+1(0+2)]|=12|[2+4+2]|=4Δzx=12211−121101=12|[2(2−0)−1(−1−1)+1(0−2)]|=12|[4+2−2]|=2∴ Area of triangle, Δ=Δxy2+Δyz2+Δzx2=22+42+22=24=26